2 Proportion Z Test Example

When dealing with proportions, such as the percentage of individuals in a sample who exhibit a certain trait or behavior, statistical tests can help determine if there’s a significant difference between groups. One of the most commonly used tests for comparing proportions is the Z-test for proportions, also known as the two-proportion Z-test. This test is particularly useful when comparing the proportions of two independent groups to see if there’s a statistically significant difference between them. Let’s dive into an example to understand how to apply the Z-test for proportions in practice.

Background

Suppose we’re interested in comparing the proportion of customers who prefer a new product feature between two different regions: the northern and southern parts of a country. We have a sample of customers from each region and want to know if the proportion of those who prefer the new feature is significantly different between the two regions.

Hypotheses

To conduct the Z-test, we first need to establish our null and alternative hypotheses:

• Null Hypothesis (H0): The proportion of customers who prefer the new feature is the same in both regions, i.e., p1 = p2.
• Alternative Hypothesis (H1): The proportion of customers who prefer the new feature is not the same in both regions, i.e., p1 ≠ p2.

Data

Let’s say our sample sizes and the number of customers who prefer the new feature in each region are as follows:

• Northern Region (Sample 1): n1 = 1000 customers, with x1 = 320 customers preferring the new feature.
• Southern Region (Sample 2): n2 = 1200 customers, with x2 = 360 customers preferring the new feature.

The proportions (p1 and p2) of customers who prefer the new feature in each region are calculated as follows:

• p1 (Northern Region) = x1 / n1 = 320 / 1000 = 0.32
• p2 (Southern Region) = x2 / n2 = 360 / 1200 = 0.30

Z-Test Formula

The formula for the Z-test statistic when comparing two proportions is:

[ Z = \frac{p_1 - p_2}{\sqrt{p_c(1-p_c)\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}} ]

Where: - ( p_c = \frac{x_1 + x_2}{n_1 + n_2} ) is the pooled proportion, combining the data from both samples to get an overall proportion of successes.

First, calculate ( p_c ):

[ p_c = \frac{320 + 360}{1000 + 1200} = \frac{680}{2200} = 0.3091 ]

Then, calculate the Z-test statistic:

[ Z = \frac{0.32 - 0.30}{\sqrt{0.3091(1-0.3091)\left(\frac{1}{1000} + \frac{1}{1200}\right)}} ]

[ Z = \frac{0.02}{\sqrt{0.3091 \times 0.6909 \times (0.001 + 0.0008333)}} ]

[ Z = \frac{0.02}{\sqrt{0.3091 \times 0.6909 \times 0.0018333}} ]

[ Z = \frac{0.02}{\sqrt{0.000309491}} ]

[ Z = \frac{0.02}{0.01757} ]

[ Z \approx 1.137 ]

Conclusion

With a calculated Z-score, we can compare it to the critical Z-score from the standard normal distribution (Z-table) for a given significance level (commonly 0.05) to determine if the difference between the proportions is statistically significant. However, given the nature of this problem and without specifying a one-tailed or two-tailed test, we’ll consider a general approach:

• For a two-tailed test at the 0.05 significance level, the critical Z-scores are approximately ±1.96.
• Our calculated Z-score (approximately 1.137) is less than 1.96, suggesting that we fail to reject the null hypothesis.

Therefore, based on this Z-test, we conclude that there is no statistically significant difference at the 0.05 significance level in the proportions of customers who prefer the new product feature between the northern and southern regions.

Practical Application

In practical terms, this means that the observed difference in preference for the new feature (0.32 in the north vs. 0.30 in the south) could be due to chance, and there might not be a real difference in the population proportions. Businesses can use this information to decide whether the observed difference is substantial enough to warrant different marketing strategies for each region.

Limitations

It’s essential to consider the limitations of the Z-test for proportions, including the assumption that the sample sizes are sufficiently large to use the normal approximation to the binomial distribution. Typically, this means that both np and n(1-p) should be greater than 5 for each sample. In our example, these conditions are met, but in cases where sample sizes are small, alternative tests like the Fisher exact test might be more appropriate.

In conclusion, the Z-test for proportions provides a powerful tool for comparing the proportions of two independent groups. By following the steps outlined and considering the assumptions and limitations of the test, researchers and business analysts can make informed decisions about whether observed differences are statistically significant, guiding further research or strategic planning.